Chapter 7, Problem 73P

To determine

The direction and speed of the acrobats after they grab each other.

Answer to Problem 73P

The final velocity is $0.64\text{m/s}$ directed $73°$ with the horizontal.

Explanation of Solution

The speed of the acrobat with mass $60\text{kg}$ is $3.0\text{m/s}$ at an angle $10°$ above the horizontal. The speed of the acrobat with mass $80\text{kg}$ is $2.0\text{m/s}$ at an angle $20°$ above the horizontal.

Since finally the two acrobats grab each other, the collision is perfectly elastic and the final velocity of both the acrobats will be same.

Write the law of conservation of linear momentum of the acrobats in x-direction.

$m_1{v}_{i1x}+m_2{v}_{i2x}=v_{fx}\left(m_1+m_2\right)$

Here, $m_1$ is the mass of the first acrobat, $v_{i1x}$ is the initial velocity of $m_1$ in x-direction, $m_2$ is the mass of the second acrobat, $v_{i2x}$ is the initial velocity of $m_2$ in x-direction, $v_{fx}$ is the final velocity in x-direction.

Re-write the above equation to get an expression for $v_{fx}$.

$v_{fx}=\frac{m_1{v}_{i1x}+m_2{v}_{i2x}}{m_1+m_2}$ (I)

Write the law of conservation of linear momentum of the acrobats in y-direction.

$m_1{v}_{i1y}+m_2{v}_{i2y}=v_{fy}\left(m_1+m_2\right)$

Here, $v_{i1y}$ is the initial velocity of $m_1$ in y-direction, $v_{i2y}$ is the initial velocity of $m_2$ in y-direction, $v_{fy}$ is the final velocity in y-direction.

Re-write the above equation to get an expression for $v_{fy}$.

$v_{fy}=\frac{m_1{v}_{i1y}+m_2{v}_{i2y}}{m_1+m_2}$ (II)

Write the formula for the magnitude of the final velocity.

$v=\sqrt{v_{fx}^2+v_{fy}^2}$ (III)

Here, $v$ is the magnitude of the final velocity.

Write the formula for the direction of the final velocity.

$\theta ={\tan }^{-1}\left(\frac{v_{fy}}{v_{fx}}\right)$ (IV)

Here, $\theta$ is the angle made by the final velocity with the horizontal.

Conclusion:

Substitute $60\text{kg}$ for $m_1$, $80\text{kg}$ for $m_2$, $\left(3.0\text{m/s}\right)\left(\cos 10°\right)$ for $v_{i1x}$, $\left(2.0\text{m/s}\right)\left(\cos 160°\right)$ for $v_{i2x}$ in equation (I).

$\begin{array}{c}{v}_{fx}=\frac{\left(60\text{kg}\right)\left(3.0\text{m/s}\right)\left(\cos 10°\right)+\left(80\text{kg}\right)\left(2.0\text{m/s}\right)\left(\cos 160°\right)}{60\text{kg}+80\text{kg}}\\ =0.192\text{m/s}\end{array}$

Substitute $60\text{kg}$ for $m_1$, $80\text{kg}$ for $m_2$, $\left(3.0\text{m/s}\right)\left(\sin 10°\right)$ for $v_{i1y}$, $\left(2.0\text{m/s}\right)\left(\sin 160°\right)$ for $v_{i2y}$ in equation (II).

$\begin{array}{c}{v}_{fy}=\frac{\left(60\text{kg}\right)\left(3.0\text{m/s}\right)\left(\sin 10°\right)+\left(80\text{kg}\right)\left(2.0\text{m/s}\right)\left(\sin 160°\right)}{60\text{kg}+80\text{kg}}\\ =0.614\text{m/s}\end{array}$

Substitute $0.192\text{m/s}$ for $v_{fx}$, $0.614\text{m/s}$ for $v_{fy}$ in equation (III).

$\begin{array}{c}v=\sqrt{{\left(0.192\text{m/s}\right)}^2+{\left(0.614\text{m/s}\right)}^2}\\ =0.64\text{m/s}\end{array}$

Substitute $0.192\text{m/s}$ for $v_{fx}$, $0.614\text{m/s}$ for $v_{fy}$ in equation (IV).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{0.614\text{m/s}}{0.192\text{m/s}}\right)\\ =73°\end{array}$

The final velocity is $0.64\text{m/s}$ directed $73°$ with the horizontal.